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aniked [119]
3 years ago
12

What is 5⁶ written in expanded form? o 5+5+5+5+5+5 5.5.5.5.5.5 ​

Mathematics
1 answer:
lara [203]3 years ago
4 0

Answer:

5x5x5x5x5x5

Step-by-step explanation:

Your multiplying 5, six times and the answer is 15,625

Hope this helepd

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Which statement is true?<br> GRAPHA<br> GRAPH B
WARRIOR [948]

Answer:

Graph B is correct

Step-by-step explanation:

5 0
3 years ago
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Please help me with this algebra PLEEASSEee!!!
amid [387]
11. 5x^2-30=0
+30 to both sides
5x^2=30
÷5 both sides
x^2=6
square root both sides
x= +square root 6 ,- square root 6

12. 4x^2+10=26
-10 both sides
4x^2=16
÷4 both sides
x^2=4
square root both sides
x=+2,-2

13. a=72yrd^2
l=2w
w=?
a=l×w
72=2w×w
72=2w^2
÷2 both sides
36=w^2
square root both sides
w=+6,-6
a measurement can't be - so w=6
Plug into l=2w
l=2×6

l=12
w=6
5 0
3 years ago
you measure the period of a mass oscillating on a vertical spring ten times as follows: period (s): 1.06, 1.31, 1.28, 0.99, 1.48
lidiya [134]

The mean and (sample) standard deviation σ = 0.2098.

<h3>What exactly would the standard deviation indicate?</h3>

The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

<h3>According to given information:</h3>

The mean is the product of the dataset's total and the sample size. Mathematically.

\bar{x}=\frac{\sum X_i}{N}

The individual periods are Xi.

The sample size is N.

\sum X i = 1.06 + 1.31 + 1.28 + 0.99,+  1.48 + 1.37+  0.98 + 1.31 + 1.59 + 1.55

\sum X i = 12.92

N = 10

While substituting the value we get:

x = 12.96/10

x = 1.292

The samples' average is 1.292.

The standard deviation:

\sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{N}}

\sum(x-\bar{x})^2 = (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.

\sum(x-\bar{x})^2 = 0.43996

Putting into the formula we get:

\sigma=\sqrt{\frac{0.43996}{10}}

σ = √(0.043996)

σ = 0.2098

The mean and (sample) standard deviation σ = 0.2098.

To know more about standard deviation visit:

brainly.com/question/18521100

#SPJ4

I understand that the question you are looking for is:

You measure the period of a mass oscillating on a vertical spring ten times as follows:

Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55

Required:

What are the mean and (sample) standard deviation?

a. Mean: 1.228, Standard Deviation: 0.2135

b. Mean: 1.325, Standard Deviation: 0.1674

c. Mean: 1.292. Standard Deviation: 0.2211

d. Mean: 1.228, Standard Deviation: 0.2098

e. Mean: 1.292, Standard Deviation: 0.2135

7 0
2 years ago
Which of the following is the area of the special trapezoid if AB = 10, CD = 10, and the height is 8.
Lelu [443]
Area of a trapezoid is A=height over 2 B1 +B2, so 10+10=20, then 20(8)=160, 160/2=80. Answer is b.
3 0
3 years ago
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If xt = 3y -1 and us = 28, find the value of y
olga55 [171]
3y - 1 = 28
+ 1 + 1
----------------
3y = 29
---- ----
3 3
y= 9.66
8 0
3 years ago
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